Problem 2 - Discrete Mathematics with Applications

Question

Find the coefficient of each. \(x^{4} y^{6}\) in the expansion of \((x-y)^{10}\)

Step-by-Step Solution

Answer

The coefficient of \(x^4y^6\) in the expansion of \((x-y)^{10}\) is 210.

Step 1: Identify the term we are looking for

We are looking for the term \(x^4y^6\). According to the binomial theorem, the term will be in the form: \[\binom{n}{k} x^{n-k} y^k\] We can now identify that n = 10. Since we want the exponent of x to be 4, and the exponent of y to be 6, we can set n-k=4 and k=6.

Step 2: Calculate the value of k

Now that we know the values of n and k, we can calculate the k value for the term we are looking for: \begin{align*} n - k &= 4 \\ 10 - k &= 4 \\ k &= 6 \end{align*} So for our term, k equals 6.

Step 3: Find the binomial coefficient

We can now compute the binomial coefficient, which is given by the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] Plugging in our values for n and k, we have: \[\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!}\] Calculate the result: \[\binom{10}{6} = \frac{10!}{6!4!} = \frac{3,628,800}{720\cdot 24} = \frac{3,628,800}{17,280} = 210\]

Step 4: Multiply the coefficients

Now, according to the binomial theorem and using the expansion we found previously: \[\binom{n}{k} x^{n-k} y^k = \binom{10}{6} x^{10-6}(-y)^6\] Substitute the values calculated: \[210 \cdot x^4 (-y)^6\] Simplify the term: \[210 \cdot x^4 y^6\] So the term we are looking for is \(210x^4y^6\) and the coefficient of the term is 210.

Key Concepts

Binomial Coefficient

When working with the Binomial Theorem, the concept of binomial coefficients becomes essential. A binomial coefficient, often denoted as \(\binom{n}{k}\), represents the number of ways to choose \(k\) elements from a set of \(n\) elements without caring about order. This concept derives from the field of combinatorics.

The binomial coefficient can be calculated using the formula:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

Here, \(n!\) (n factorial) represents the product of all positive integers up to \(n\). The factorial function grows quickly, which can make calculation by hand tedious for large \(n\) and \(k\).

Using the binomial coefficient allows us to expand expressions raised to a power, such as \((x-y)^{10}\), and find specific terms within that expansion. It helps simplify complex polynomial expressions by providing a systematic way to calculate coefficients of individual terms.

Polynomial Expansion

Polynomial expansion refers to the process of expanding a binomial expression raised to a power using the Binomial Theorem. This theorem provides a formula for expressing \((x + y)^n\) as a sum of terms involving binomial coefficients.

The essential formula for binomial expansion is:

• \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k\)

Each term in the expansion is of the form \(\binom{n}{k} x^{n-k} y^k\). In every term, the exponents \(n-k\) and \(k\) will sum up to \(n\), ensuring that each expansion respects the original degree of the polynomial.

This method is incredibly powerful for breaking down expressions like \((x-y)^{10}\) into a sum of more manageable terms. It is a vital tool when dealing with algebraic problems that require identifying specific terms within an expanded polynomial form.

Combinatorial Mathematics

Combinatorial mathematics is an area of mathematics concerned with counting, arranging, and analyzing discrete structures. It's all about finding how many ways something can be done, and it's essential for understanding the principles behind the binomial theorem.

In the context of binomial expansion, combinatorial mathematics studies the number of ways to distribute items. This directly relates to the binomial coefficient, since it quantifies the different ways to choose a subset from a larger set without considering order.

For example, when solving problems like finding the coefficient of \(x^4y^6\) in \((x-y)^{10}\), we model the different possible combinations of \(x\) and \(y\) in each of the expanded terms using the principles from combinatorics. This allows us to understand not just how to calculate a result but why the calculation works, providing a deeper insight into the nature of algebraic structures.